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modified 6 years ago

Is this how a class-H amp works

6

98

217

02:12:29

If so, how is it done?

published 8 years ago

8 years ago

👍

8 years ago

Very crude, but yeah, that's how class G amp would work, not a class H amp. A class H amp modulates the supply rail continuously with the output voltage, not only when a given output level is reached. The difference is that class G amps mainly switch between two or a few more supply voltages to satisfy the output requirements, while a class H amps has an infinite amount of supply voltage states, dictated solely by the output waveform, e.g. the supply rail follows the output signal's curves, but just a few volts above the output voltage (much like a bootstrap works). This enables the main amp (let's say class A) to work at a constant dissipation (efficiency) while maintaining good linearity, while the rest of the power requirement is shifted to a more efficient amp - like a class B, or even a class D switching amp, although I haven't seen a class D representation of the circuit.

8 years ago

Basically a class H amp can be viewed as a bootstrap using nonlinear components - like transistors. A class G amp, just switches between a few voltage levels to handle the load requirements

8 years ago

It works a little bit like this http://everycircuit.com/circuit/4870765004718080 and is most often used to boost the efficiency of a class A amp, so notice I changed the bias slightly

8 years ago

Did you post it? You know I can't look at circuits unless their posted. By the way, I didn't look at the link because it's a pain to copy it by hand. Also, @hurz, stay out of this. All you might end up doing is getting him to just do one of his retreat techniques where nothing useful can happen. @thebugger, if the supply rail modulation is controlled by a class-B amp, then how is it done? You know that unless things like inductors are used, the amount of current going in is the amount of current going out, in which case the efficiency is limited by the supply, which has to be higher than the peak voltage of the wave. So, do a little math and try to prove me wrong. There is no convincing me unless you show me some math.

8 years ago

Jason, im sick of you! What you are doing is clear, buggzy is just an idiot fraud. Is this what you tell us? Leave him alone, he will not change. Jason, you are far away from having long experience in electronic, but you are much more intellegent then buggzy. Its clear buggzy is again wrong and cant see that. Let him be stubborn, you do not change anything. Keep on learning, but not on the price of buggzy destruction. You are both in the same situation, but YOU try to distroy him. Hands off you little evil.

8 years ago

http://everycircuit.com/circuit/6279323151761408

8 years ago

If you know how a bootstrap works, it's basically the same thing, although designed with nonlinear components utilising different amp classes

8 years ago

Check out the example. It's a crude version of such an amp. The output stage works in class A mode (the active devices working continuously, in their linear region). The supply rails are being driven by a more efficient amp (class B), with a feedback from the output of the class A amp. This way the class B amp always swings 5V above the output of the class A amp. As you know constant voltage multiplied by constant current gives a constant power. The class A amp always dissipates the same amount of power no matter the required power at the output. For instance if the class A amp is biased at 20W dissipation, even if the output is 200W the class A amp would still dissipate 20W. The difference will be handled by the more efficient class B amp. The idea behind this is very evident. You can clearly see the crossover distortion from the class B amp, they're efficient, but prone to higher distortion levels, whereas class A amps, are very inefficient, but also very linear. A class H amp provides the best of both worlds, an efficiency close to that of a class B amp, and linearity close to that of a class A amp. This is somewhat different than class AB amps, where the amp is forced to work in class B mode above a given output level, where here the amp is always working in class A mode, just using the efficiency of a class B amp (at least close to it)

8 years ago

But you still have to factor in the efficiency losses of the class-B amp. What did I say about showing me some math? A class-A amp only has such high fidelity because it consumes much more current than necessary. A class-B amp gets its efficiency because it uses exactly the amount of current needed. Where is the extra current for the class-A amp going to come from? It can't come from the class-B amp because what makes a class-B amp class-B is the fact that it doesn't consume extra current. The amount of current going out is equal to the amount of current going in. Everyone knows this! If the class-A amp is using extra current to make it class-A, then where is that current going to come from? It could come from a class-D amp because it uses uses inductors to make extra current to make up for the voltage drop, but it can't come from a class-B amp which never contains inductors. Just make a simple class-H amplifier, measure the average current going in, multiply that by the power supply voltage to get wattage consumed, then measure the average current going out, multiply that by the average output voltage to get output wattage, then divide the output wattage by the input wattage and show me the number you get. That's all I ask for, a simple class-H power amplifier, and a little math to determine efficiency. @hurz, you need to leave him alone because you trying to teach him something will degrade into name-calling, which will further degrade into silence, and that won't get you anywhere! And, you've already failed at teaching him this at least once, and you've already failed how to teach him other things many times. Let me try to teach him using my own methods. Maybe he'll understand, and maybe he won't. Also, I am NOT trying to destroy him. I'm just using my own approach, and, if he somehow finds it to harsh, then let him just ignore me.

8 years ago

Could you please post that circuit you gave the link to? I guess I could technically copy that circuit from the image, but there's no way to find the component values. Also, I know that you can't post that circuit directly because it's derived, but you can make a copy.

8 years ago

I just copied and pasted that circuit with the image and guessed the component values by trial and error, and I fail to see how it is any more efficient by adding another layer of transistors. All that means to me is that there's more transistors that the current has to pass through more transistors. Just do a little math and see what efficiency your class-H amp has. I haven't measured it myself, and maybe I'm wrong and it really does have some amazing efficiency, but you have to show me what efficiency number you got and the math you used to get it.

8 years ago

Okay, If you don't get the circuit, what does math of all things do you good? I really can't dumb it down more.

8 years ago

And also, this is a very crude circuit, it won't do you much good like so, it needs refining

8 years ago

http://everycircuit.com/circuit/6279323151761408

8 years ago

"And also, this is a very crude circuit, it won't do you much good like so, it needs refining" SHUT UP BUGGZY!

8 years ago

I'm pretty sure you just gave me the same link. Maybe you should make a good class-H power amplifier like your ultimate hi-fi amplifiers, and measure the efficiency. It's dumb just using theory and blind faith trusting that your amp is efficient. You should measure it yourself to make sure and prove @hurz wrong! It will also give some interesting information like just how efficient it is, and maybe you can experiment and create a highly efficient class-H amp. I'm sure it'll be fun and interesting.

8 years ago

Yeah it'd be, but lately I've been very busy. It takes time refining a circuit. And such an amp is actually two circuits combined into 1.

8 years ago

It's just exemplary, and yeah it's the same link. It's unlisted, so you should be able to open it

8 years ago

Class H amps aren't as efficient as you may think though. The advantage is that they draw a smaller idle current than normal class A amps, at any given output level. A class A amp may dissipate a 100W, for only 1W of output power, whereas a class H amp would dissipate let's say 35W+1W at idle. The efficiency is around 50-60%

8 years ago

I'm confused. I thought what makes a class-A amp class-A is it's high idle current. If it has a lower idle current, then wouldn't it be class-AB? If so, then what's the point of driving it through a class-B amp? Also, you could just make the amp a simple one that's driven by the simulators perfect OP-amp.

8 years ago

That's the thing, the idle current isn't lower, it's the same. The supply voltage is lower, and since power is a multiplication of voltage and current, the lower voltage decrease the power as well. So instead of 30Vx5A for instance, you have 5Vx5A. The rest of the supply voltage (the other 25V) is provided by the class B amp, which is more efficient.

8 years ago

30x5=150 ... 5x5=25 + 5x25=125 = 150! Were is the advantage? Or better shut up.

8 years ago

@hurz is right, except for that last sentence. @hurz, if we all just go quiet on this topic, then he won't be able to learn anything, and he'll keep spreading his false ideas. @thebugger, could you please make a simple class-H amp that uses EC's ideal OP-amp? Since a complex amplifier should be very close to an ideal amplifier, the result shouldn't be very different, but it will be much easier to make. Could you just do this one simple favor for me? This could be a chance to completely prove me and @hurz wrong!

8 years ago

http://everycircuit.com/circuit/6470257197973504

8 years ago

He's not right, 5x5 is only viable for the class A amp, the class B part behaves in a different way (there's no idle current constantly dissipating). So for a class A amp it would be 30x5=150W even at 25W output power, and for the class H amp it would be 5x5=25W plus 44W from the class B amp (ideally). Calculating the efficiency at 25W from these numbers we get 16% efficiency for the class A amp and 36% efficiency for the class H amp, which further increases with output power until it reaches about 50-60% efficiency

8 years ago

Basically you can view it as this. The idle point of the class A amplifier is the zero current point of the class B amp. Every curve above or below the idle current of the class A amp, gets followed by the class B amp, but since it doesn't have any idle current, it doesn't dissipate as much heat. The class A amp is being kept under constant potential, while the class B amp only fills in the blanks

8 years ago

What you seem to be describing is a class-AB amp with the A and the B separated into two output stages. Also, I found an online simulator that includes vacuum tubes! It's a bit harder to use than EC, and is for a desktop, not a mobile (although it could be used for mobile, it'll just be very awkward). It's literally called "circuit simulator", and has a few advantages over EC, one of which is that it has many more components including many IC's, polarized capacitors, JFETs, and even triodes. It has no obvious space limit, although I haven't played with it enough to know for sure.

8 years ago

Also, why do you simply refuse to make a class-H amp?

8 years ago

How many times do I need to send you different links to class H amps I made?

8 years ago

It's not my fault you have an iPhone and are too lazy to write the links manually.

8 years ago

http://everycircuit.com/circuit/4535821107724288

8 years ago

The two links you gave me were exactly the same. I'll take a look at this one.

8 years ago

You haven't posted it. How many times have I told you I can't view circuits that you haven't posted?

8 years ago

It's unlisted not private, you should be able to view it

8 years ago

I'll post it

8 years ago

http://everycircuit.com/circuit/4535821107724288

8 years ago

I can't go to links except via the internet, except I can't view the circuit there either. It needs to be public. Thanks for posting it. I might be busy for a couple of hours though, so don't expect an immediate response.

8 years ago

After much, much, much, frustration involving modifying your amp and then deciding to make my own version, but getting nothing but crashes, crashes, and more crashes, I decided I would try to measure the efficiency of your amp as it is. I got 40% efficiency, which is much higher than the theoretical efficiency of 25% for a class-A amplifier that has the pushing part of the output stage as a constant current source. I would still like to see if you can manage one that is higher than 50% in efficiency, although this is a good start. Let's see if @hurz can find some excuse this time. The only problem I can think of is that because of the way you set it up, the class-B part of the amp might still cause crossover distortion at the output. I still don't see why one would prefer this kind of amp over other another kind of amp, especially class-G. I personally don't think that a class-A or H amp is worth the trouble of lower efficiency for reduced distortion, because I can hardly see how I could notice 0.1% distortion, or even 1% distortion, although I do admit that I've never heard an audio signal with 1% distortion, because to me all this simulator is, and probably will ever be, is just a game.

8 years ago

This type of amplifier is most suitable at high power levels, where you'd want a class A fidelity, and a class AB (or close enough) efficiency. For instance a normal class A amp at 250W and a 25% efficiency, would dissipate 1kW of power, even at idle, which is highly unacceptable. A class H amp would dissipate (at idle) let's say 40W instead of 1kW. When throttled up, the class H amp would dissipate 40W + whatever efficiency your class B amp can give you. Anyway the class A part of the amp has a considerably high ripple rejection, so crossover distortion from the class B amp, would not occur on the output.

8 years ago

Btw the human ear is variably sensitive to distortion. A 1% for you may be ok, but to me could be totally unacceptable. Anyway, since the class A part of the amp always operates at a low enough power, and the class B amp's distortion doesn't influence the output, the class H amp would virtually have the same distortion at 1W and at 250W for instance. A 0.1% distortion at 250W is veeery good.

8 years ago

Ok, I see, but the class-B amp seems to be controlling the ground of the class-A amp allowing the low voltage power supply of the class-A amp power a high voltage load. I'd imagine that the crossover distortion of the class-B amp would still go through the class-A amp because the class-A amp is in series with the class-B amp, and the class-A amp would need to have a very quick response to negate the crossover distortion since it can easily be filtered out using a lowpass filter and thus must be made up of very high frequency harmonics.

8 years ago

Well, no.

8 years ago

Why?

8 years ago

Because the class A amp sees nothing more than a constant voltage supply rail. As far as it's concerned there's no such rail modulation, there's 5V at 5A and that's it. And, no the class A amp needn't have such a high response rate because it's the one controlling the class B circuit not the other way around. I repeat again the class A amp has a given ripple rejection that rejects any power supply made noise.

8 years ago

Ok.

8 years ago

Where's @hurz? Can he disprove this amp?

8 years ago

Idk idc :d

8 years ago

Que diablos paso aqui!!! 😲

8 years ago

The class A implementation from Mr fary tales buggzy does nothing here, its completely over ruled by the class B amp, as you guys declared the right opamp it is. So what we have here is a pure class "B" with an resistor presented by a not working class A dead circuit. By the way, a class B amp is in best case for high power <75% in efficency. Better you calc with 65% and for low power calc much much below <<65%.

8 years ago

Do not apply more than 1Vpp to buggly class A amp which is sooooo perfect. http://everycircuit.com/circuit/5593534147330048 and over engineeeeeeered as a graduated engineer should know better. Shut up. Please shut up.

8 years ago

I don't see your point here, as in you don't have a point, you just have to be a total douche - as always.

8 years ago

Did you make it public? Apparently @thebugger was a bit confused about what my definition of "posted" meant. I hope you aren't also confused. In case you are, unlisted does not work as "posted" for me, it needs to be public.

8 years ago

Sorry, I didn't make it clear. In the first sentence, "it" refers to the link.

8 years ago

Jason9 already told mister fary-tales what the point of an efficent class H amp MUST be, a quick buck converter which must be able to work follow audio frequencies. And surprise this buck converter is a Class D amp and not a shitty class B or AB. Think about "class D -> fast and efficent buck converter".

8 years ago

Efficiency is improved right? So is linearity. What sort of amp does that make it then? According to Elliott Sounds class H and G amps have a very broad definition and can be viewed as ANY amp that uses rail modulation

8 years ago

Your class A amp is riding on a "class B" amp. The B amp makes the noisy virtual ground for the A amp. Ground noisy supply noisy, how can the class A amp reject this noise if he does not even see it? He will just follow it with a rejection of 0dB. BTW put a realistic input impedance e.g. 10kOhm? And stop talking about profis in this business, we are talking what YOU have invented and present here. 2 weeks ago you not even know the difference about distortion and noise. Instead of coming with an over complicated amp concept, try to keep it simple like jason did in the past. You shows up with this kind of circuit to blind yourself and the community.

8 years ago

Lousy ripple rejection http://everycircuit.com/circuit/5337267138985984

8 years ago

Look at my circuit and make sure to read the description. The class-A amp sees the noise because it's reference voltage is noisy but the speaker's reference voltage isn't which causes a varying speaker current, and since a class-A amp without feedback is current based and not voltage based, the class-A amp will remove this noise because all it is is a constant current source going in series with a voltage controlled current source, and since on top is a constant current source and the current of the bottom current source is controlled only by the input voltage, the current going out will be very stable and because a speaker is mostly a resistor, it produces a very stable output.

8 years ago

How many circuits shell I analyse in parallel for this respectfull community full of ignorand foul mouths?

8 years ago

Why have you choosen such a high current? To bring a swing of plus minus 5V against 4Ohm the current source can be 1.25A while the driver BJT must be able to sink two times 1.25A so 2.5A http://everycircuit.com/circuit/5856954692141056 next replace the perfect current source with a BJT real implementatiin and make some noise/ripple on this class A amplifier while the input is off.

8 years ago

So jason, again take this 16digit number 5856954692141056 and enter it HERE in EveryCircuit search field manual. If you dont want to spend some money for better tools (android device or EC for google chrome for PC) and if its to anoying for you to enter 16digits to follow a discussion, let me know and we can stop it.

8 years ago

http://everycircuit.com/circuit/4518931719258112 still a pretty decent ripple rejection

8 years ago

Anyrhing you have to analysis in my circuits directly, or you still come with circuits to blind yourself and the community. The class A amp is riding on class B amp. All noise is given to the loudspeaker. Efficency is class B minus class A. Read again Rod Elliot carefully and listen to "switched modulated power supply" and not "modulated supply".

8 years ago

@hurz oh my God! That's amazing! Now I can finally view unlisted circuits properly. Also, it is such a high current because it has to be able to supply the 4ohm load with up to 50V because the current is based not on the supply voltage but on the output voltage. The supply voltage is lower than the output voltage so that there can be a high ripple rejection class-A in there without it using up so much power. The output can be such a high current and still work because the supply voltage is boosted by the class-AB amp, which is at a high efficiency because it doesn't boost the two supplies separately, but they're paired together to give a theoretically non-existent idle current for the class-AB to supply so that it can function like normal.

8 years ago

Read the description of my circuit again. In case you somehow don't know which one, its number is 6456608630767616.

8 years ago

12.5A^2 x 4Ohm = 625Watt hey ho.

8 years ago

You missed a step. (12.5*2^(1/2))^2=12.5^2/2. Therefore, the appropriate equation is 12.5^2/2*4=312.5W out. This is at maximum output voltage which would be impossible due to losses, and therefore the output power consumed by the class-AB amplifier (which is assumed to be pure class-B) would be 1.28 (inverse of 78.5%) times more input power (400W). The class-A amplifier adds an additional amount of input power, which can be found by multiplying the input voltage (5V) by 12.5A, and then multiplying that by two because there are two supplies gives 125W, which added to the 400W gives 525W, which means 60% efficiency. It's probably 50% efficiency at max output level because there was some amount of rounding in my equations, and more importantly, there are voltage losses across the transistors limiting the maximum output power.

8 years ago

And you missed a lot http://everycircuit.com/circuit/5856954692141056

8 years ago

Start first to understand class A before coming to class H. Your class A here http://everycircuit.com/circuit/5517568658112512

8 years ago

Okay efficiency is not class B minus class A, because that means the more efficient the class A amp is, the less efficient the overall amp is. You probably mean an arithmetic medium between the two amps? Then an approximate maximum theoretical efficiency would be 64% which is still pretty awesome. And as you see in my example the class A amp still has a pretty decent ripple rejection.

8 years ago

What I mean is, the class A amp just reduce the class B efficiency. And its all in all just a Class B amp, cuz the ripple rejection is not given as you say. 1kOhm input impedance, and were is the output impedance from the audio source e.g. MP3 player, CD player? Read again what Rod Elliot wrote to class H & G and you will understand how the supply must look like to have an efficient amplifier. What you invented here is a class B amp with on top a wasteful class A amp. Remove the A amp and you have better efficiency and equal THD.

8 years ago

BTW, what is an "arithmetic medium"? I guess you mean "arithmetic mean"? What ever, an average, unweighted is just a new mix and is still not precise math. As long you guys dont see whats wrong with this topology it doesn't make sense to start exact calculations. Go ten steps back and keep circuits as basic as possible.

8 years ago

Ok here's your input impedance http://everycircuit.com/circuit/5528729533284352

8 years ago

Cool, I was expecting you are coming with a 10kOhm version but now we are even a step ahead and can already lift your class A amp to e.g. 10V virtual ground. Check that http://everycircuit.com/circuit/5424245327003648

8 years ago

Yeah, but that's not a real representation of what the circuit does. The supply rail is modulated with the same voltage as the output. Basically the virtual ground and the output never become differential at 10V. The two op amps in the original schematic make sure of that.

8 years ago

How do they do that and how does then the loudspeaker get a delta in voltage?

8 years ago

Next week I will be on vacation and far away in the caribic with bad internet connection. Hurry up.

8 years ago

EC doesn't require much, just a few kbps. Anyway I'm telling you what I'm seeing on the simulation. One end of the speaker is grounded (to the real ground), the other gets 200-300uV of signal.

8 years ago

One end of the loudspeaker grounded. The other side we lift to 10V and what do you see then?

8 years ago

It's not the other side you lift. The virtual ground and the output are meant to be lift simultaneously.

8 years ago

What????

8 years ago

Sure the output lifts cause the class B amp lifts the virtual ground. And if this happens the input transistor of the class A amp gets some extra current which will hit the A amp like a punchball. Any noise from the B amp does cause strange reaction of the A amp. Which is suppressed by the B amp while the B is not perfect the distortion is going directly or even strange amplified by the A amp on the loudspeaker. Test it yourself with an AC source as replacement for the B amp WITHOUT feedback, connect the AC to virtual ground and play with DC and AC.

8 years ago

Again, I took just your lastest class A part with offset conpensating opamp. Removed both ripple ac source, better placed only one to represent the class B amp. Play with DC and AC of virtual ground. http://everycircuit.com/circuit/5054972628828160

8 years ago

What you need to avoid this effect is a galvanic separation, but I dont think a transformer is not what a highend simple class H suppose to be. One at input and one at output?

8 years ago

The class-A amp's output and its virtual ground always being at the same voltage seems counterintuitive, but it is to keep it within its operating region with its low voltage supplies. What the class-A amp sees is the ground for the speaker moving up and down, and it tries to keep its output at 0V (in relation to its virtual ground) thus causing current to move through the speaker. Of course, it's actually the other way around, the class-A amp tries to push some current through the speaker, and the class-AB amp lifts its virtual ground such that the class-A amp sees the ground of the speaker going up and down so that the class-A amp will always have approximately 0V (in relation to its virtual ground) at its output in order to supply the right amount of current, since all the class-A amp is is just two current sources, one feeding into the other such that it acts as a varying current source, not caring what its output voltage is, only caring what its output current is.

8 years ago

Jason, you just talk about a vision and not about this implementation. Please test the circuits we publish before swagger about a well know highside amp concept. Practical work is importante

8 years ago

The more you say about this problem and that problem and now that it won't work the way I describe, it just seems more and more as if you never even saw my class-H amp circuit! Read its description carefully, look at the simulation to see that it works properly, look at the voltage across the constant current source to see that it is always positive so that it will work if made with transistors, and then see if any of your "problems" are still valid! I didn't just come up with those distortion values in the circuit, I actually measured them using a 500uH/500uF tank circuit to produce an input (that's been separated by an OP-amp to keep the tank circuit from losing energy), and a twin-T notch filter (using two 1kOhm resistors, one 500ohm resistor, two 500nF capacitors, and one 1uF capacitor) to filter out the output, and since they both match perfectly in frequency, the only signals going out are harmonics, and those aren't diminished at all because the twin-T notch filter is very precise.

8 years ago

How interessting

8 years ago

Don't give me that. Tell me what problems are still valid and if you still think the class-A amp has a low ripple rejection, even though it improves the output distortion so dramatically.

8 years ago

Jason, im in hurry, anyway buggzy is still messing with his implementation, you have problems with basic links to follow, and I spend/waste to much time now. I saw your latest V2.0 version and you still use a transformer. Is this a high fidellity audio Amplifier?? Galvanic separation is the key to do such "highside" concepts as I told buggzy, but not for a high end low distortion one, were you spend 12A for a class A Amp in idle. You talk about ripple rejection of the A amp, but its not there. It even does give us gain from any ripple we have on one of the separated 5V power supplies. Buggzys idea is there but it does not work. Or would you like to use optocoupler to be ground free with all its nonlinearity? Transformers are already bad enough.... but the ripple rejection fary-tales is proved to be wrong. Here again: http://everycircuit.com/circuit/5347828597784576

8 years ago

Transformer coupled amps can be made quite good, with high fidelity and everything, but a high end audio transformer is very expensive and not worth the cost in most cases. Only tube amps exclusively use output transformers and even they can be made transformerless, but that's a waste of tubes and power for me

8 years ago

How interessting

8 years ago

When are you planning on coming back from your vacation?

8 years ago

Lets see

8 years ago

Do you mean that you are just going to come back when you feel like it meaning that you could be gone anywhere from two days to two years?

8 years ago

Right, more or less.

8 years ago

Ok.

8 years ago

Back again ;-)

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